Question: Simplify the following expression: $\dfrac{120n^3}{70n^5}$ You can assume $n \neq 0$.
Solution: $ \dfrac{120n^3}{70n^5} = \dfrac{120}{70} \cdot \dfrac{n^3}{n^5} $ To simplify $\frac{120}{70}$ , find the greatest common factor (GCD) of $120$ and $70$ $120 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5$ $70 = 2 \cdot 5 \cdot 7$ $ \mbox{GCD}(120, 70) = 2 \cdot 5 = 10 $ $ \dfrac{120}{70} \cdot \dfrac{n^3}{n^5} = \dfrac{10 \cdot 12}{10 \cdot 7} \cdot \dfrac{n^3}{n^5} $ $\phantom{ \dfrac{120}{70} \cdot \dfrac{3}{5}} = \dfrac{12}{7} \cdot \dfrac{n^3}{n^5} $ $ \dfrac{n^3}{n^5} = \dfrac{n \cdot n \cdot n}{n \cdot n \cdot n \cdot n \cdot n} = \dfrac{1}{n^2} $ $ \dfrac{12}{7} \cdot \dfrac{1}{n^2} = \dfrac{12}{7n^2} $